Problem: $su - 8t + 3u + 1 = 9t + 6u + 4$ Solve for $s$.
Solution: Combine constant terms on the right. $su - 8t + 3u + {1} = 9t + 6u + {4}$ $su - 8t + 3u = 9t + 6u + {3}$ Combine $u$ terms on the right. $su - 8t + {3u} = 9t + {6u} + 3$ $su - 8t = 9t + {3u} + 3$ Combine $t$ terms on the right. $su - {8t} = {9t} + 3u + 3$ $su = {17t} + 3u + 3$ Isolate $s$ $s{u} = 17t + 3u + 3$ $s = \dfrac{ 17t + 3u + 3 }{ {u} }$